Question

For g.p. if a=3,r=2/3 then S6=?

Answer 1

answer : 665/81

explanation : first term of the G.P, a = 3

common ratio , r = 2/3

using formula,

sum of n terms , Sn = a(1 - rⁿ)/(1 - r) , when 0 < r < 1

here, n = 6 , a = 3 and r = 2/3

so, S6 = 3[1 - (2/3)^6 ]/(1 - 2/3)

= 3(1 - 64/729)/(1/3)

= 3(729 - 64)/(729/3)

= (3 × 665)/(243)

= (665)/(81)

hence, sum of 6 terms = 665/81

Answer 2

$\bf{\underline{\underline{Answer:}}}$

$\bf{\approx 1.095}$

$\bf{\underline{\underline{Step\: by\: step\: explanation:}}}$

∵ Sum of a GP is,

$S_n=\frac{a(r^n-1)}{r-1}$ ( when r > 1 )

$S_n=\frac{a(1-r^n)}{1-r}$ ( when 0 < r < 1 )

Where,

a = first term,

r = common ratio,

n = number of terms,

Here, a = 2, r = $-\frac{2}{3}$ n = 6,

$\implies S_6=\frac{2(1-(-\frac{2}{3})^6)}{1+\frac{2}{3}}$

=1.09465020576

$\approx 1.095$